POJ 2135 费用流基础

题意：给出N个点，M条边，问从1-N来回走一次最短路径是多少，且一条边只能经过一次。

直接一遍费用流即可。不过题目中初值需要注意，贡献了几次WA。

[cpp]
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <stack>
#include <map>
#include <iomanip>
#define PI acos(-1.0)
#define Max 100005
#define inf 2000000000//一开始inf = 1 << 28 但是WA，所以注意一下，这个初值得大。
#define LL(x) (x<<1)
#define RR(x) (x<<1|1)
#define REP(i,s,t) for(int i=(s);i<=(t);++i)
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define mp(a,b) make_pair(a,b)
using namespace std;
inline void readint(int &ret)
{
    char c;
    do
    {
        c = getchar();
    }
    while(c < '0' || c > '9');
    ret = c - '0';
    while((c=getchar()) >= '0' && c <= '9')
        ret = ret * 10 + ( c - '0' );
}
int n , m ;
struct kdq
{
    int s ,e ,l , c ,next ;
}ed[1000005] ;
int head[10005] ,num ;
void add(int s ,int e ,int l ,int c)
{
    ed[num].s = s ;
    ed[num].e = e ;
    ed[num].l = l ;
    ed[num].c = c ;
    ed[num].next = head[s] ;
    head[s] = num ++ ;
    ed[num].s = e ;
    ed[num].e = s ;
    ed[num].l = 0 ;
    ed[num].c = -c ;
    ed[num].next = head[e] ;
    head[e] = num ++ ;
}
int S , T ;
void init()
{
    mem(head,-1) ;
    num = 0 ;
}
int dis[Max] ;
bool vis[Max] ;
int qe[Max * 10] ;
int pre[Max] ,path[Max] ;
int spfa()
{
    REP(i,0,T)dis[i] = inf ,vis[i] = 0 ;
    dis[S] = 0 ;
    vis[S] = 1 ;
    int h = 0 , t = 0 ;
    qe[h ++ ] = S ;
    while( h > t )
    {
        int tt = qe[t ++ ] ;
        vis[tt] = 0 ;
        for (int i = head[tt] ;~i ;i = ed[i].next )
        {
            int ee = ed[i].e ;
            int l = ed[i].l ;
            int c = ed[i].c ;
            if(l > 0 && dis[ee] > dis[tt] + c)
            {
                pre[ee] = tt ;
                path[ee] = i ;
                dis[ee] = dis[tt] + c ;
                if(!vis[ee])
                {
                    vis[ee] = 1 ;
                    qe[h ++ ] = ee ;
                }
            }
        }
    }
    return dis[T] != inf ;
}
int MFMC()
{
    int M = 0 ;
    while(spfa())
    {
        int ff = inf ;
        for (int i = T ;i != S ;i = pre[i])
        ff = min(ff , ed[path[i]].l) ;
        for (int i = T ;i != S ;i = pre[i])
        ed[path[i]].l -= ff ,ed[path[i] ^ 1].l += ff ;
        M += dis[T] * ff ;
    }
    return M ;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("acm.txt", "r", stdin);
#endif
    init() ;
    readint(n) ;
    readint(m) ;
    S = 0 ,T = n + 1 ;
    REP(i,0,m - 1)
    {
        int a , b , c ;
        readint(a) ;
        readint(b) ;
        readint(c) ;
        add(a,b,1,c) ;
        add(b,a,1,c) ;
    }
    add(S,1,2,0) ;
    add(n,T,2,0) ;
    cout << MFMC() << endl;
    return 0;
}

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <stack>
#include <map>
#include <iomanip>
#define PI acos(-1.0)
#define Max 100005
#define inf 2000000000//一开始inf = 1 << 28 但是WA，所以注意一下，这个初值得大。
#define LL(x) (x<<1)
#define RR(x) (x<<1|1)
#define REP(i,s,t) for(int i=(s);i<=(t);++i)
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define mp(a,b) make_pair(a,b)
using namespace std;
inline void readint(int &ret)

补充：软件开发 , C++ ,